JEE Main 2024Inverse Trigonometric FunctionsInverse Trigonometric FunctionsHardQuestionFor α,β,γ≠0\alpha, \beta, \gamma \neq 0α,β,γ=0, if sin−1α+sin−1β+sin−1γ=π\sin ^{-1} \alpha+\sin ^{-1} \beta+\sin ^{-1} \gamma=\pisin−1α+sin−1β+sin−1γ=π and (α+β+γ)(α−γ+β)=3αβ(\alpha+\beta+\gamma)(\alpha-\gamma+\beta)=3 \alpha \beta(α+β+γ)(α−γ+β)=3αβ, then γ\gammaγ equalsOptionsA3\sqrt{3}3B32\frac{\sqrt{3}}{2}23C12\frac{1}{\sqrt{2}}21D3−122\frac{\sqrt{3}-1}{2 \sqrt{2}}223−1Check AnswerHide SolutionSolutionLet sin−1α=A,sin−1β=B,sin−1γ=C\sin ^{-1} \alpha=A, \sin ^{-1} \beta=B, \sin ^{-1} \gamma=Csin−1α=A,sin−1β=B,sin−1γ=C A+B+C=π(α+β)2−γ2=3αβα2+β2−γ2=αβα2+β2−γ22αβ=12⇒cosC=12sinC=γcosC=1−γ2=12γ=32\begin{aligned} & \mathrm{A}+\mathrm{B}+\mathrm{C}=\pi \\ & (\alpha+\beta)^2-\gamma^2=3 \alpha \beta \\ & \alpha^2+\beta^2-\gamma^2=\alpha \beta \\ & \frac{\alpha^2+\beta^2-\gamma^2}{2 \alpha \beta}=\frac{1}{2} \\ & \Rightarrow \cos \mathrm{C}=\frac{1}{2} \\ & \sin \mathrm{C}=\gamma \\ & \cos \mathrm{C}=\sqrt{1-\gamma^2}=\frac{1}{2} \\ & \gamma=\frac{\sqrt{3}}{2} \end{aligned}A+B+C=π(α+β)2−γ2=3αβα2+β2−γ2=αβ2αβα2+β2−γ2=21⇒cosC=21sinC=γcosC=1−γ2=21γ=23